Home Right Triangles Angles in Standard Position Sine Law & Ambiguous Case Cosine Law Problem Solving Summary&Test

UNIT 6  : BASIC TRIGONOMETRY WITH TRIANGLES

LESSON 2:  ANGLES IN STANDARD POSITION HOMEWORK QUESTIONS   (Solutions below)

Definition:  An angle is in standard position if it has its vertex at the origin and  initial arm along the positive x-axis.  The terminal arm is found by rotating the initial arm about the origin to a terminal position in one of the 4 quadrants.  The rotation is positive if it is in the counter – clockwise direction and negative if in the clockwise direction.

To find the trigonometric ratios of angles between 90o and 180o, use the following rules:

2.  Evaluate each of the following using reference (related acute) angles.  Give answers to 5 decimal places.

a)  cos 121o                              b)  sin 106o                               c) cot 97o                                 d) sin 165.4o

3.  Evaluate each of the following directly using your calculator.  Give answers to 5 decimal places.

a)  sin 1190                               b) tan 143.6o                            c)  sec 124.9o                           d) sin 148.2o

a)  sin B = 0.3256                    b)  cos B = -0.87921               c)  csc B = 1.5894                   d) cos B = 0.3685

Solutions:

Solution:

 x = -6 y = 8 r = 10

2.  Evaluate each of the following using reference (related acute) angles.  Give answers to 5 decimal places.

a)  cos 121o                              b)  sin 106o                               c) cot 97o                                 d) sin 165.4o

Solutions: [Recall to get the reference (related acute) angle, subtract from 180o].

a)         cos 121o  = - cos (180o – 121o)

= - cos 59o                            ** 59o is the reference (related acute) angle

= - 0.51504                          ** Using your cacluator

b)         sin 106o   = sin (180o – 106o)

= sin 74o                               ** 74o is the reference (related acute) angle

= 0.96126                            ** Using your cacluator

b)      sin 165.4o   = sin (180o – 165.4o)

= sin 14.6o                            ** 14.6o is the reference (related acute) angle

= 0.25207                            ** Using your cacluator

3.  Evaluate each of the following directly using your calculator.  Give answers to 5 decimal places.

a)  sin 1190                               b) tan 143.6o                            c)  cos 124.9o                           d) sin 148.2o

Solutions:

a)  sin B = 0.3256                    b)  cos B = -0.87921               c)  csc B = 1.5894                   d) cos B = 0.3685

Solutions:

a) Since sin B is positive, < B could be in either the first or second quadrant.  There are two solutions.

b) Since cos B is negative, < B will be in  the second quadrant.

c) Since csc B is positive, < B could be in either the first or second quadrant.  There are two solutions.

d) Since cos B is positive, < B will be in  the first quadrant.